Blog Log in Get Started Free

Physics MCQs for NEET — Practice Questions with Answers

Q: Are these Physics questions free to attempt?

Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

Q: Do I need an account to practise Physics MCQs?

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

Q: Are these previous year NEET questions?

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.

Practice free Physics NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Chapter Language Difficulty Show only questions I answered incorrectly Register free to filter questions

● ● ● ● ●

We consider a thermodynamic system. If ΔU represents the increase in its internal energy and W the work done by the system, which of the following statements is true ?

ΔU = –W in an adiabatic process

ΔU = W in an isothermal process

ΔU = –W in an isothermal process

ΔU = W in an adiabatic process

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

According to the first law of thermodynamics

$Δ Q = Δ U + Δ W$

In adiabatic process $Δ Q = 0$ , hence $Δ U = - Δ W$

● ● ● ● ●

The volume of a gas is reduced adiabatically to $\frac{1}{4}$ of its volume at 27°C, if the value of γ = 1.4, then the new temperature will be -

350 × 4^0.4K

300 × 4^0.4 K

150 × 4^0.4 K

None of these

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

For adiabatic change $T V^{γ - 1}$ = constant

⇒ $\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1}$ ⇒ $T_{2} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} \times T_{1}$

⇒ $T_{2} = {(\frac{V}{V / 4})}^{1 .4 - 1} \times 300 = 300 \times {(4)}^{0 .4} K$

● ● ● ● ●

For an adiabatic expansion of a perfect gas, the value of $\frac{Δ P}{P}$ is equal to

$- \sqrt{γ} \frac{Δ V}{V}$

$- \frac{Δ V}{V}$

$- γ \frac{Δ V}{V}$

$- γ^{2} \frac{Δ V}{V}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$P V^{γ} =$ constant : Differentiating both sides

$P γ V^{γ - 1} d V + V^{γ} d P = 0 \Rightarrow \frac{d P}{P} = - γ \frac{d V}{V}$

● ● ● ● ●

A gas expands under constant pressure P from volume V₁ toV₂. The work done by the gas is

$P (V_{2} - V_{1})$

$P (V_{1} - V_{2})$

$P (V_{1}^{γ} - V_{2}^{γ})$

$P \frac{V_{1} V_{2}}{V_{2} - V_{1}}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Work done $= P Δ V = P (V_{2} - V_{1})$

● ● ● ● ●

When heat in given to a gas in an isobaric process, then

The work is done by the gas

Internal energy of the gas increases

Both (1) and (2)

None from (1) and (2)

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

When heat is supplied at constant pressure, a part of it goes in the expansion of gas and remaining part is used to increase the temperature of the gas which in turn increases the internal energy.

● ● ● ● ●

One mole of a perfect gas in a cylinder fitted with a piston has a pressure P, volume V and temperature 273 K. If the temperature is increased by 1 K keeping pressure constant, the increase in volume is

$\frac{2 V}{273}$

$\frac{V}{91}$

$\frac{V}{273}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

For isobaric process $\frac{V_{2}}{V_{1}} = \frac{T_{2}}{T_{1}} \Rightarrow V_{2} = V \times \frac{274}{273}$

Increase $= \frac{274 V}{273} - V = \frac{V}{273}$

● ● ● ● ●

Unit mass of a liquid with volume V₁ is completely changed into a gas of volume V₂ at a constant external pressure P and temperature T. If the latent heat of evaporation for the given mass is L, then the increase in the internal energy of the system is -

Zero

$P (V_{2} - V_{1})$

$L - P (V_{2} - V_{1})$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$Δ Q = Δ U + P Δ V$

⇒ mL = ΔU + P(V₂ – V₁)

⇒ ΔU = L – P (V₂ – V₁) (âˆµ m = 1)

● ● ● ● ●

A gas expands 0.25m³ at constant pressure 10³ N/m², the work done is -

2.5 ergs

250 J

250 W

250 N

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$Δ W = P Δ V = 10^{3} \times 0 .25 = 250 J$

● ● ● ● ●

If 300 ml of a gas at 27°C is cooled to 7°C at constant pressure, then its final volume will be -

540 ml

350 ml

280 ml

135 ml

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

V ∝ T at constant pressure

⇒ $\frac{V_{1}}{V_{2}} = \frac{T_{1}}{T_{2}}$

⇒ $V_{2} = \frac{V_{1} T_{2}}{T_{1}} = \frac{300 \times 280}{300} = 280 m l .$

● ● ● ● ●

Which of the following is correct in terms of increasing work done for the same initial volume and final volume ?

Adiabatic < Isothermal < Isobaric

Isobaric < Adiabatic < Isothermal

Adiabatic < Isobaric < Isothermal

None of these

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.